PDE for Heat Diffusion Equation

[ZedCar]

Homework Statement

The one-dimensional heat diffusion equation is given by :

∂t(x,t)/∂t = α[∂^2T(x,t) / ∂x^2]

where α is positive.

Is the following a possible solution? Assume that the constants a and b can take any positive value.

T(x,t) = exp(at)cos(bx)

Homework Equations

The Attempt at a Solution

T(x,t) = exp(at)cos(bx)

∂T/∂t = a exp(at) cos(bx)
∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

As a, b and α are all positive this cannot be a solution.


A friend and I were working on this and got the answer above. Though as it was about a week ago I can't exactly remember what we've done here.

In the first step I think we've differentiated exp(at) with respect to t, treating the cos(bx) as a constant multiplier, which become a exp(at) cos(bx).

Is that all correct above?

Then we differentiated again with respect to t.

So wouldn't that become;

∂^2 T/∂x^2 = a^2 exp(at) cos(bx)

Why did we previously get ∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

Thanks!

 

[HallsofIvy]

Homework Statement

The one-dimensional heat diffusion equation is given by :

∂t(x,t)/∂t = α[∂^2T(x,t) / ∂x^2]

That first "t" should be "T". I realize this is a typo.

where α is positive.

Is the following a possible solution? Assume that the constants a and b can take any positive value.

T(x,t) = exp(at)cos(bx)

Homework Equations

The Attempt at a Solution

T(x,t) = exp(at)cos(bx)

∂T/∂t = a exp(at) cos(bx)
∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

As a, b and α are all positive this cannot be a solution.


A friend and I were working on this and got the answer above. Though as it was about a week ago I can't exactly remember what we've done here.

In the first step I think we've differentiated exp(at) with respect to t, treating the cos(bx) as a constant multiplier, which become a exp(at) cos(bx).

Is that all correct above?

Then we differentiated again with respect to t.

Why?

So wouldn't that become;

∂^2 T/∂x^2 = a^2 exp(at) cos(bx)

Yes, but the second derivative with respect to t is irrelevant. Your differential equation has only the first derivative with respect to t.

Why did we previously get ∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

Because that is the correct second derivative with respect to x, not t.

Thanks!

The first derivative with respect to t is a exp(at()cos(bx). If the differential equation ∂T/∂t= a∂^2T/∂x^2 were satisfied by exp(at)cos(bx) you would have to have
aexp(at)cos(bx)= -ab^2exp(at)cos(bt) which would imply a= -ab^2 which in turn gives 1= -b^2 so b^2= -1 which is impossible for real valued b.

 

[ZedCar]

Cheers for that! I see what I was doing wrong - getting x and t confused...

 

[ZedCar]

Another possible solution we investigated for the same heat diffusion equation was;

T(x,t) = exp(-at + ibx)The PDEs being;

∂T/∂t = -a exp(-at + ibx)

∂^2 T / ∂x^2 = - b^2 exp(-at + ibx)So the possible solution is possible if a = b^2 α
I understand all the above now.

Though apparently T(x, t) is complex in this instance. How is this apparent?

 

[Ray Vickson]

Homework Statement

The one-dimensional heat diffusion equation is given by :

∂t(x,t)/∂t = α[∂^2T(x,t) / ∂x^2]

where α is positive.

Is the following a possible solution? Assume that the constants a and b can take any positive value.

T(x,t) = exp(at)cos(bx)

Homework Equations

The Attempt at a Solution

T(x,t) = exp(at)cos(bx)

∂T/∂t = a exp(at) cos(bx)
∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

As a, b and α are all positive this cannot be a solution.


A friend and I were working on this and got the answer above. Though as it was about a week ago I can't exactly remember what we've done here.

In the first step I think we've differentiated exp(at) with respect to t, treating the cos(bx) as a constant multiplier, which become a exp(at) cos(bx).

Is that all correct above?

Then we differentiated again with respect to t.

So wouldn't that become;

∂^2 T/∂x^2 = a^2 exp(at) cos(bx)

Why did we previously get ∂^2 T/∂x^2 = -b^2 exp(at) cos(bx)

Thanks!

Be very glad that a, b > 0 does not give a solution, since such a "solution" would violate some basic laws of Physics: warm regions (where cos(bx) > 0) would get hotter, and cool regions (where cos(bx) < 0) would get colder, so heat would flow from cold to hot without any external influence---violating Thermodynamics. Also, warm regions would eventually get hotter than the hottest star, and cold regions would plunge well below absolute zero.

Look instead, at the case a < 0, b > 0. Does that behave properly?

RGV